Why is the group $Z_6$ under addition isomorphic to the group $Z_7^*$ under multiplication?
So I'm trying to answer this question:
Q. Which of the set are isomorphic to each other?
$S_3$, $Z_6$, $Z_3 \times Z_2$ and $Z_7^*$
Now I know that $S_3$ is not because its not abelian while $Z_6$, $Z_3 \times Z_2$ are since my professor said this:
The group $Z_n$ under addition is an abelian group which means $Z_6$ is abelian.
Now another idea is:
The direct product of two cyclic groups $Z_m$ and $Z_n$ is isomorphic to ($Z_{mn}$,+) iff m and n are relatively prime.
For $Z_3 \times Z_2$, 3 and 2 are relatively prime which means $Z_6$ and $Z_3 \times Z_2$ are isomorphic.
Now, whats throwing me off is $Z_7^*$. I know $Z_7^*$ is composed of the elements {1,2,3,4,5,6}. So i've researched a bit online, and $Z_7^*$ has an order of 6 which I'm not understanding.
Also, it was mentioned that $Z_7^*$ is abelian. Therefore, are all groups $Z_n^*$ under multiplication an abelian group? and why does $Z_7^*$ have an order of 6? Also, I've seen people say $Z_7^*$ is cyclic and has a generator such as 3, I'm not understanding that concept or in other words, how is 3 a generator? because if i do $3^6$=729 and that is not divisible by 7.
Or if anyone has a different way to do this problem, that would help as well. I'm open to any ideas. thanks!
Best Answer
Let's look at the structure of $(\Bbb Z_7)^{\times}$ in some more detail.
$[1]$ isn't very interesting, it's clearly the (multiplicative) identity, though.
So now let's look at $[2]$, specifically, its powers.
$[2]^2 = [2]\cdot [2] = [4]$. This is...boring.
$[2]^3 = [2]\cdot[2]\cdot[2] = [8] = [1]$ (because $8 = 1 + 7$, so $8 \equiv 1$ (mod $7$)).
This tells us $[2]$ has order $3$.
OK, so now let's look at $[3]$.
$[3]^2 = [9] = [2]$. Thus $[3]^6 = ([3]^2)^3 = [2]^3 = [1]$. So the order of $[3]$ divides $6$ (this is, of course, self-evident by Lagrange).
As we saw above $[3]^2 = [2] \neq [1]$, so the order of $[3]$ is NOT $2$.
$[3]^3 = [27] = [6] \neq [1]$, so the order of $[3]$ is NOT $3$.
This means the order of $[3]$ must be $6$, so it generates the entire group:
$[3]^1 = [3]$
$[3]^2 = [2]$
$[3]^3 = [6]$
$[3]^4 = [4]$ ($81 = 4 + 7\ast 11$)
$[3]^5 = [5]$ ($243 = 5 + 7\ast 34$)
$[3]^6 = [1]$
Therefore, an isomorphism between $(\Bbb Z_6,+)$ and $((\Bbb Z_7)^{\times},\ast)$ is:
$[k]_6 \mapsto [3^k]_7$ (where the subscripts/brackets mean "equivalence class modulo").
Note $[3]^{-1} = [5]$, so we could have used $[5]$ as a generator instead.